Question

HELP! I will give medal and fan!

How do you find the roots of x^4 + 3x^2 - 4 = 0?

Answer 1

Well the first thing you should notice is that the only powers of \(x\) are 2 and 4. Both of which are even. So let's do a substitution for \(y=x^2\). Then your equation is the same as \[y^2+3y-4=0.\]Can you find the roots of this?

Answer 2

Okay, I see how you got that but I don't even know how to find the roots of that. Like what's the first step?

Answer 3

Alright. To factor quadratics, you need to try and find two numbers \(a\) and \(b\) such that \(ab=-4\) and \(a+b=3\). If you find these, then we have\[(y+a)(y+b)=y^2+(a+b)y+ab=y^2+3y-4\]so we factored the quadratic.

Answer 4

Alright.

Answer 5

So can you tell me what values I might have for \(a\) and \(b\)? Feel free to take a minute to think about it if you need to.

Answer 6

ay + by + ab =?

Answer 7

Where did you get that equation from? All I'm looking for are two integers \(a\) and \(b\) so that \(ab=-4\) and \(a+b=3\).

Answer 8

Okay then how did you get that? @KingGeorge

Answer 9

That comes from experience having done these problems before/other people telling me to this :P

Answer 10

I just don't understand how you got any of that. @KingGeorge

Answer 11

Have you been taught the FOIL method of multiplying binomials?

Answer 12

Is that the first, outside, inside, last thing?

Answer 13

yes. That's how I got that \[(y+a)(y+b)=y^2+(a+b)y+ab\]

Answer 14

Ohh alright.

Answer 15

So you see that if we find some integers \(a\) and \(b\), then we can solve for \(y\).